1. Tardigrade
  2. Question
  3. Mathematics
  4. If (1+x)15=a0+a1 x+ ldots+a15 x15, then displaystyle∑r=115 r (ar/ar-1) is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $(1+x)^{15}=a_{0}+a_{1} x+\ldots+a_{15} x^{15}, $ then $\displaystyle\sum_{r=1}^{15} r \frac{a_{r}}{a_{r}-1}$ is equal to

EAMCETEAMCET 2005

Solution:

Given that
$(1+x)^{15}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{15} x^{15}$
$\Rightarrow { }^{15} C_{0}+{ }^{15} C_{1} x+{ }^{15} C_{2} x^{2}+\ldots+{ }^{15} C_{15} x^{15}$
$=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{15} x^{15}$
Equating the coefficient of various powers of $x$, we get
$a_{0}={ }^{15} C_{0}, a_{1}={ }^{15} C_{1}, a_{2}={ }^{15} C_{2}, \ldots, a_{15}={ }^{15} C_{15}$
$\therefore \displaystyle\sum_{r=1}^{15} r \frac{a_{r}}{a_{r-1}}=\sum_{r=1}^{15} r \frac{{ }^{15} C_{r}}{{ }^{15} C_{r-1}}$
$= \displaystyle\sum_{r=1}^{15} r \frac{\frac{15 !}{r !(15-r) !}}{\frac{15 !}{(r-1) !(15-r+1) !}}$
$= \displaystyle\sum_{r=1}^{15} \frac{(r-1) !(15-r+1) !}{r !(15-r) !}$
$= \displaystyle\sum_{r=1}^{15} 15-r+1=15+14+13+\ldots+2+1$
$=\frac{15(15+1)}{2}=120$